# How do I solve for Capacitance in this circuit?

I know that the equivalent resistance of this circuit will be

$$R_1-j/(\omega*C_x)+L_1+(R_2-C_1)||L_2$$ $$(3+.814j)-j/(4737*C_x)+[(4j)(30-8j)]/(30-4j)$$

I don't know what to do from here; I could set the top equation equal to Z, but Z is not given. I think I have to use the power factor given, but I have no idea what to do with it. Can someone tell me how power factor comes into all of this?

• The total impedance is $3\:\Omega+j\:814\:\text{m}\Omega+\left(j\:4\:\Omega\mid\mid\left(30\:\Omega -j\:8\:\Omega\right)\right)+C_x$. Just work out the first part without $C_x$ and express it as a complex number, $Z$. If $\operatorname{Im}\left(Z\right)>0$, then the addition of a capacitor can achieve the goal. Just solve for $-j\:\operatorname{Im}\left(Z\right)=\frac{1}{j\:\omega\:C}$. (You are told $\omega$, already.) This is because the imaginary part has to go to zero in this case.
– jonk
May 5 '18 at 23:48